{"id":1275,"date":"2018-01-17T05:52:43","date_gmt":"2018-01-17T05:52:43","guid":{"rendered":"http:\/\/punjabirevolution.com\/?page_id=1275"},"modified":"2018-01-17T05:53:18","modified_gmt":"2018-01-17T05:53:18","slug":"bayes-theorem-examples","status":"publish","type":"page","link":"http:\/\/punjabirevolution.com\/index.php\/bayes-theorem-examples\/","title":{"rendered":"Bayes\u2019 Theorem Examples"},"content":{"rendered":"

Bayes’ Theorem Examples<\/u><\/strong><\/h1>\n

Example 1<\/u><\/strong><\/h2>\n


\nHistorical Data.<\/u><\/strong><\/p>\n

P(cavity) = 0.1                    — 10%<\/p>\n

P(toothache) = 0.05         — 5%<\/p>\n

 <\/p>\n

P(Cavity | toothache) = 0.8   — 80%<\/p>\n

 <\/p>\n

 <\/p>\n

Future Prediction: <\/u><\/strong><\/p>\n

P(toothache | cavity)  = ?<\/p>\n

 <\/p>\n

using bays law à P(B|A) = {P(A|B)P(B)}\/P(A)<\/p>\n

P(toothache | cavity)  = {P(cavity | toothache) P(toothache) } \/ P(cavity)<\/p>\n

=             (0.8*0.05)\/0.1 = 0.4<\/p>\n

Example 2
\n<\/u><\/strong><\/h2>\n

If HIV has a prevalence of 3% in San Francisco, and a particular HIV test has a false positive rate of .001 and a false negative rate of .01, what is the probability that a random person who tests positive is actually infected (also known as “positive predictive value”)?<\/p>\n

Solution of Example 2<\/h3>\n

We have to find the probability of hiv+ given the report is positive.<\/p>\n

Using Bayes theorem
\nP(hiv+|t+) = P(t+|hiv+)P(hiv+) \/ P(t+)<\/p>\n

Now,<\/p>\n

false positive  P(t+|hiv-) = .001<\/p>\n

P(t-|hiv-) =1-.001 =  .999<\/p>\n

false negative = P(t-|hiv+) = .01
\nP(t+|hiv+) = 1-.01 = .99<\/p>\n

P(t+) = P(t+&hiv+) + P(t+&hiv-)
\n=P(t+|hiv+)P(hiv+) + P(t+|hiv-)P(hiv-)
\n= .99*.03 + .001*.97<\/p>\n

Putting all these values in equation
\nP(hiv+|t+) = P(t+|hiv+)P(hiv+) \/ P(t+)<\/p>\n

= (.99 * .03)\/{.99*.03 + .001*.97}<\/p>\n

= .968   ===> 96.8%<\/p>\n

Bayes’s Theorem Example 1<\/p>\n